A body of radius `R` and mass `m` is rolling smoothly with speed `v` on a horizontal surface. It then rolls up a hill to a maximum height `h`. If `h = 3 v^2//4g`. What might the body be ?

Options

A. a solid circular cylinder
B. a hollow circular cylinder
C. a solid circular sphere
D. a hollow circular sphere.

Answer

A

Explanation

(a) Interpreting `h` as the height increase for the centre of mass of the body, then mechanical energy conservation, `K_i = U_f`, leads to
`(1)/(2) mv_(com)^2 + (1)/(2) I omega^2 = mgh rArr (1)/(2) mv^2 + (1)/(2) I ((v)/(R))^2`
=`mg ((3v^2)/(4 g))`
from which `v` cancels and we obtain `I = (1)/(2) mR^2`
This moment on inertia is the moment of inertia solid cylinder.